If $$S = \left\{x \in [0, 2\pi] : \begin{vmatrix} 0 & \cos x & -\sin x \\ \sin x & 0 & \cos x \\ \cos x & \sin x & 0 \end{vmatrix} = 0 \right\}$$, then $$\displaystyle\sum_{x \in S} \tan\left(\frac{\pi}{3} + x\right)$$ is equal to:

We first note that the determinant condition in the set $$S$$ is

$$\begin{vmatrix} 0 & \cos x & -\sin x \\[4pt] \sin x & 0 & \cos x \\[4pt] \cos x & \sin x & 0 \end{vmatrix}=0.$$

We expand this determinant along the first row. Using the usual $$3\times3$$ cofactor rule

$$\begin{vmatrix} a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{vmatrix} =\;a_{11}(a_{22}a_{33}-a_{23}a_{32})-a_{12}(a_{21}a_{33}-a_{23}a_{31})+a_{13}(a_{21}a_{32}-a_{22}a_{31}),$$

we substitute

$$a_{11}=0,\;a_{12}=\cos x,\;a_{13}=-\sin x,$$ $$a_{21}=\sin x,\;a_{22}=0,\;a_{23}=\cos x,$$ $$a_{31}=\cos x,\;a_{32}=\sin x,\;a_{33}=0.$$

So we have

$$\begin{aligned} \Delta(x)&=0\Bigl(0\cdot0-\cos x\cdot\sin x\Bigr) \\ &\qquad-\,\cos x\Bigl(\sin x\cdot0-\cos x\cdot\cos x\Bigr) \\ &\qquad+\;(-\sin x)\Bigl(\sin x\cdot\sin x-0\cdot\cos x\Bigr). \end{aligned}$$

The first term is clearly zero. In the second term the bracket becomes

$$\sin x\cdot0-\cos x\cdot\cos x=0-\cos^2x=-\cos^2x,$$

so the whole term is

$$-\cos x\,(-\cos^2x)=\cos x\,\cos^2x=\cos^3x.$$

In the third term the bracket is

$$\sin x\cdot\sin x-0\cdot\cos x=\sin^2x-0=\sin^2x,$$

hence the term contributes

$$(-\sin x)\,\sin^2x=-\sin^3x.$$

Collecting everything,

$$\Delta(x)=\cos^3x-\sin^3x.$$

The determinant being zero means

$$\cos^3x-\sin^3x=0.$$

For real numbers the cube function is one-one, so

$$\cos^3x=\sin^3x\;\Longrightarrow\;\cos x=\sin x.$$

Dividing by $$\cos x$$ (which is allowed because $$\cos x=0$$ would force $$\sin x=0$$ making both zero, impossible for a valid point on the unit circle), we get

$$\tan x=1.$$

Now the general solution of $$\tan x=1$$ is

$$x=\frac{\pi}{4}+n\pi,\qquad n\in\mathbb Z.$$

Restricting to the given interval $$[0,2\pi]$$ we have

$$x_1=\frac{\pi}{4},\qquad x_2=\frac{5\pi}{4}.$$

Thus $$S=\left\{\frac{\pi}{4},\;\frac{5\pi}{4}\right\}.$$

We must now evaluate the sum

$$\sum_{x\in S}\tan\!\left(\frac{\pi}{3}+x\right) =\tan\!\left(\frac{\pi}{3}+\frac{\pi}{4}\right)+\tan\!\left(\frac{\pi}{3}+\frac{5\pi}{4}\right).$$

We first compute $$\tan\!\left(\frac{\pi}{3}+\frac{\pi}{4}\right).$$ We state the tangent-addition formula:

$$\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\,\tan B}.$$

With $$A=\frac{\pi}{3}$$ and $$B=\frac{\pi}{4}$$ we substitute $$\tan\frac{\pi}{3}=\sqrt3$$ and $$\tan\frac{\pi}{4}=1:$$

$$\tan\!\left(\frac{\pi}{3}+\frac{\pi}{4}\right)= \frac{\sqrt3+1}{1-\sqrt3\cdot1}=\frac{\sqrt3+1}{1-\sqrt3}.$$

To simplify the fraction we multiply numerator and denominator by the conjugate $$1+\sqrt3:$$

$$\frac{\sqrt3+1}{1-\sqrt3}\cdot\frac{1+\sqrt3}{1+\sqrt3} =\frac{(\sqrt3+1)(1+\sqrt3)}{1-3} =\frac{\sqrt3+3+1+\sqrt3}{-2} =\frac{4+2\sqrt3}{-2} =-2-\sqrt3.$$

So

$$\tan\!\left(\frac{\pi}{3}+\frac{\pi}{4}\right)=-2-\sqrt3.$$

Next we tackle $$\tan\!\left(\frac{\pi}{3}+\frac{5\pi}{4}\right).$$ Observe that

$$\frac{\pi}{3}+\frac{5\pi}{4}=\frac{4\pi}{12}+\frac{15\pi}{12}=\frac{19\pi}{12}.$$ Since $$\tan(\theta+\pi)=\tan\theta,$$ we write $$\frac{19\pi}{12}=\pi+\frac{7\pi}{12},$$ and therefore

$$\tan\!\left(\frac{19\pi}{12}\right)=\tan\!\left(\frac{7\pi}{12}\right).$$

But we have just found $$\tan\!\left(\frac{7\pi}{12}\right)=-2-\sqrt3.$$ Hence

$$\tan\!\left(\frac{\pi}{3}+\frac{5\pi}{4}\right)=-2-\sqrt3.$$

Adding the two values, we obtain

$$(-2-\sqrt3)+(-2-\sqrt3)=-4-2\sqrt3.$$

Hence, the correct answer is Option B.